Chemical calculations

Transcription

1 Chemical calculations Stoichiometry refers to the quantities of material which react according to a balanced chemical equation. Compounds are formed when atoms combine in fixed proportions. E.g. 2Mg + O 2 2MgO 2 atoms Mg + 2 atoms oxygen 2 molecules MgO (x4) 8 atoms Mg + 8 atoms oxygen 8 molecules MgO (x6) 1 dozen atoms Mg + 1 dozen atoms oxygen 1 dozen molecules MgO BUT atoms and molecules are tiny, so counting them in ones and twos is silly. Instead we count atoms and molecules in moles. A mole is x units. Count up the number of atoms in exactly 12g of pure 12 C. That number is a mole. Class Exercise A mole (a) is 6 x units, and (b) weighs the atomic weight (for atoms) or molecular weight (for molecules) in grams. We can find the atomic weight (same as atomic mass or relative atomic mass) from the periodic table, and molecular weights by adding atomic masses. E.g. Atomic weights in g: Na 23, Cl 35.5, C 12, O 16 What is the weight of (i) 3 moles of NaCl? (ii) 0.2 moles O 2? (iii) 0.2 moles of O atoms? (iv) 11 moles CO 2? How many molecules are there in those quantities? How many atoms? 1

2 2Mg + O 2 2MgO How do we know what will happen when magnesium and oxygen react? We know the reactants, so we write down their chemical symbols remembering that oxygen is a diatomic molecule. Writing O for oxygen would imply that the reaction involves atomic oxygen whose properties are very different from that of molecular O 2 or ozone O 3 and would thus result in a serious error. Finding the formula for the product is often more challenging. Experimentally, we could determine the empirical formula by weighing the mass of magnesium that reacted with excess oxygen and weighing the mass of the product formed. 2Mg (s) + O 2 (g) 2MgO (s) Equations must be balanced an unbalanced equation implies creation or destruction of mass. It is common to include the physical state of reactants and products in brackets by their symbols. If an element exists as different allotropes (e.g. diamond and graphite, or white, red, violet or black phosphorus) we also indicate this, usually as a subscript e.g. C graphite (s) 2

3 Simple stoichiometry (works for most situations!) 1. Write down the equation 2. Balance it 3. If necessary, convert the known amount(s) reactant/product to moles 4. Use the balanced equation to set up the appropriate mole ratios 5. Use the mole ratios to work out the number of moles of desired reactant/product 6. If necessary, convert back to grams How many grams of oxygen are required for the complete combustion of 50 grams methane? 1. Write down the equation 2. Balance it 3. If necessary, convert the known amount(s) of reactant/product to moles 4. Use the balanced equation to set up the appropriate mole ratios 5. Use the mole ratios to work out the number of moles of desired reactant/product 6. If necessary, convert back to grams 1. CH 4 (g) + O 2 (g) CO 2 (g) + H 2 O (g) 2. CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O (g) 3. 1 mole of methane weighs { x 1} = 16 g, so 50 g is 50/16 = moles 4. From the equation 1 mole of methane reacts with two moles of oxygen 5. So the amount of oxygen required is 2 x = 6.25 moles 6. To turn this back into grams we need the molecular weight of oxygen, which is 2 x 16 = 32g (since the oxygen molecule consists of 2 atoms joined together). Thus the amount of oxygen required is 6.25 moles x 32 g/mole = 200 g. 3

4 Lithium hydroxide is used in space vehicles to absorb carbon dioxide. How much CO 2 can be absorbed by 1 kg of hydroxide? 1. Write down the equation 2. Balance it 3. If necessary, convert the known amount(s) of reactant/product to moles 4. Use the balanced equation to set up the appropriate mole ratios 5. Use the mole ratios to work out the number of moles of desired reactant/product 6. If necessary, convert back to grams 1. LiOH (s) + CO 2 (g) Li 2 CO 3 (s) + H 2 O (l) 2. 2LiOH (s) + CO 2 (g) Li 2 CO 3 (s) + H 2 O (l) 3. Molar mass of LiOH is { } = g/mol. Thus, # of moles of hydroxide is 1000/23.94 = 41.8 mol 4. From the equation, 2 moles of LiOH react with 1 mole of CO 2 5. So the amount of CO 2 that reacts is 41.8/2 = 20.9 moles 6. The MW of CO 2 is { x 16} = 44 g/mole. So the mass of CO 2 that reacts is 20.9 moles x 44 g/mole = 920g. 14g methane and 14g water react to give hydrogen and carbon monoxide. Which reactant is in excess? How much hydrogen is formed? 1. CH 4 (g) + H 2 O (g) H 2 (g) + CO (g) 1. Write down the equation 2. Balance it 3. If necessary, convert the known amount(s) of reactant/product to moles 4. Use the balanced equation to set up the appropriate mole ratios 5. Use the mole ratios to work out the number of moles of desired reactant/product 6. If necessary, convert back to grams 2. CH 4 (g) + H 2 O (g) 3H 2 (g) + CO (g) 3. MW of methane is { x 1} = 16 g/mole, so there are 14/16 = moles methane. MW of water is { x 1} = 18, so there are moles of water [3a. The equation shows that methane and water are used up at the same rate, so water is the limiting reactant - it will be consumed completely] 4. 1 mole water reacts to give 3 moles of hydrogen 5. So the # of moles of hydrogen formed is 3 x = moles 6. MW of H 2 is 2 x 1 = 2, so the weight of hydrogen formed is 2 x = g 4

5 Finding the empirical formula Suppose we started with mg magnesium and obtained mg product. What is the empirical formula? As the product only contains oxygen and magnesium any gain in mass must be due to oxygen chemically bound in the compound mg mg = mg of oxygen in the product. To get to the stoichiometric ratio of the two elements and express an empirical formula we need to convert mass to moles: 1 mole of Mg is g, so g Mg / = 8.70 x 10-3 mol 1 mole of oxygen (atoms) is 16 g, so g O is /16 = x 10-3 mol Dividing both amounts by the smaller number of moles yields the mole ratio of the two elements: x 10-3 mol / x 10-3 mol = x 10-3 mol / x 10-3 mol = Within experimental error the Mg:O ratio is 1:1 so we write the formula MgO (not Mg 1 O 1 ). The empirical formula expresses the smallest whole number ratio of atoms in a compound. There are two major kinds of stoichiometry problems you will encounter: a) You know the amount of one reactant b) You know the amount of at least two reactants E.g. a) How much magnesium oxide will form if 5.73 g magnesium burns in excess oxygen? 0) Think moles! 1) Convert mass of Mg to moles of Mg 5.73 g /24.31 g mol -1 = 2.36 x 10-1 mol Mg 2) Convert moles of Mg to moles of MgO using the coefficients in the balanced equation: 2.36 x 10-1 mol Mg 2.36 x 10-1 mol MgO 3) Convert moles of MgO to mass of MgO To find the molar mass (or formula weight) of MgO add the mass of all atoms in the formula ( g mol g mol -1 = g mol -1 ) x 10-1 mol MgO = 9.50 g MgO Keep more significant figures in intermediate calculations to avoid rounding errors. 5

6 b) A sample of 1.51 g magnesium is ignited in a sealed 1L flask that contains 1.42 g oxygen gas. How much magnesium oxide can be formed and how much of which reactant is left behind? First determine which reactant limits the amount of product: 0) Think moles! 1) Convert the masses of the reactants to moles of reactant 1.51 g Mg = 6.21 x 10-2 mol Mg 1.42 g O 2 = 4.44 x 10-2 mol O 2 2) Identify the limiting reactant The limiting reactant is the chemical that will be used up first. There are more moles of magnesium than oxygen, but the two reactants may not be consumed at the same rate. It s simplest to calculate how much product could be formed from the given amount of each reactant x 10-2 mol Mg 6.21 x 10-2 mol MgO (if enough O 2 is available) 4.44 x 10-2 mol O x 10-2 mol MgO (if enough Mg is available) Though there is more magnesium by mass and mols it is the limiting reactant, so all the magnesium will be consumed and some oxygen will be left over. 3) Calculate the mass of product that can form 6.21 x 10-2 mol MgO = 2.50 g MgO (Use MW of MgO, which is the AW of Mg and O combined) Some oxygen remains. Since we know the masses of reactants and products we can use the law of conservation of mass to calculate the mass of oxygen left over. The mass of the reactants is 1.51 g Mg g O 2 = 2.93 g reactants. Since only 2.50 g of product is formed the difference must be O 2 left over: 2.93 g 2.50 g = 0.43 g oxygen left un-reacted. 6

7 Had we mistakenly assumed that oxygen was the limiting reactant, we d calculate that 8.88 x 10-2 mol of MgO should have been produced. This amount of MgO would weigh: 8.88 x 10-2 mol MgO = 3.58 g MgO which is more than the mass of reactants available (2.93 g). An error is not always found so readily but it pays to check if your result makes sense. This example has practical relevance. Magnesium is sometimes used as a getter in tungsten light bulbs where it removes trace amounts of oxygen and nitrogen; this greatly enhances the lifetime of the light bulb (as long as you ensured that oxygen is the limiting reactant). Solution Stoichiometry In aqueous solution the number of moles of a component dissolved per litre of solution is molar concentration, or molarity of that component. This is simpler to use than other possible measures of concentration, such as g/l, %, or ppm. 7

8 A solution contains 2.4 g NaOH in 50 ml solution; what is the molarity? Molarity is moles/litre, so we need the number of moles and the volume in litres. # moles = wt/mw = 2.4 g / 40 g mol -1 = 0.06 moles Volume = L, so molarity = 0.06 mol / 0.05 L = 1.2M 23.4 ml of a M solution of NaOH was used to neutralize ml of a sulfuric acid sample of unknown concentration. (i) (ii) (iii) Write down and balance the equation for the reaction. How many moles of NaOH react? What is the molarity of the sulfuric acid? 2NaOH + H 2 SO 4 Na 2 SO 4 + 2H 2 O The number of moles of sodium hydroxide is: M V = mol/l L = mol NaOH 8

9 2NaOH + H 2 SO 4 Na 2 SO 4 + 2H 2 O At the equivalence point (neutral) the number of moles of sulfuric acid can be calculated using the coefficients in the balanced chemical equation Remember, we need two moles of base per mole of acid: mol NaOH = mol H 2 SO 4 As this is the number of moles in 0.01L, the molarity of the sulfuric acid is thus: M = / 0.01 = M 3g of ethanoic (acetic) acid is dissolved in 50 ml water. The resulting solution is neutralised by 41 ml of sodium hydroxide solution. Determine the molarity of both solutions 1. Molarity = moles/litre 2. MW of ethanoic acid, CH 3 COOH, = { x x } = 60 g/mol, so 3g = 3/60 =0.05 moles 3. Total volume of solution = 53 ml = 0.053L, so molarity of the acid is 0.05/0.053 = M 4. CH 3 COOH + NaOH CH 3 COONa + H 2 O, so 1 mole of acid reacts with 1 mole of base. 5. Therefore number of moles of NaOH = Volume of base = 41 ml, so molarity of the base = 0.05/0.041 = 1.22M 9

10 % Yield Often we need to calculate the % yield of a reaction. This is the amount of product isolated from the reaction mixture divided by the maximum theoretical amount of product, expressed as a percentage. Example: 4.75 g metallic copper is precipitated from a solution containing g CuSO 4.5H 2 O by adding excess iron metal. Calculate the % Yield. Cu 2+ (aq) + Fe(s) Cu(s) + Fe g (CuSO 4.5H 2 O) = mol CuSO 4.5H 2 O, which contains mol Cu 2+ This is the maximum amount of copper that could be formed mol Cu 2+ = 6.42 g Cu % yield = 4.75/6.42 x 100 % = 74.0 % 10